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What is the difference between your approach, and calculating 9 9 9 directly? Is one better than the other? If so, why?
It is not apparent to me that finding the accuracy of 9 9 × lo g 1 0 9 is easily done. Note that the "initial few digits" is actually determined by the "fractional part of "9^9 log 9".
I differ with your answer Agnishom Chattopadhyay
The value of
9
9
9
is
9
9
9
=
9
3
8
7
4
2
0
4
8
9
=
1
.
9
6
6
2
7
0
5
×
1
0
7
7
So the answer for your question i.e., first 3 digits will be 1 9 6 So once have a look.
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You have calculated ( 9 9 ) 9 instead.
If you simply threw 9^9^9 into a calculator, that is what it would have given you, because it works from left to right. Remember that tower of exponents is done top down (hence right to left).
How do you calculate that?
I agree with Chandra Sekhar Vadigalla
(Similar idea to Agnishom Chattopadhyay)
Let
N
=
9
9
9
=
a
1
⋯
a
n
b
1
⋯
b
m
. Observe that
a
1
⋯
a
n
⋅
1
0
m
≤
N
<
a
1
⋯
a
n
⋅
1
0
m
+
1
Then we get
lo
g
1
0
a
1
⋯
a
n
+
m
≤
lo
g
1
0
N
<
lo
g
1
0
a
1
⋯
a
n
+
m
+
1
which implies
(
fractional part of
lo
g
1
0
N
)
=
lo
g
1
0
a
1
⋯
a
n
−
(
n
−
1
)
This makes the calculation faster.
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Press [ 9 ][ * ][ = ][ = ][ = ]... or [ 9 ][ * ][ * ][ = ][ = ][ = ].. depends onto type of calculator we are using, observe changes of first 3 digits.
There is valuable theory being identified within this week. To tell a hint, first few digits belong to propagating figures while last few digits belong to no squeezed out requirement. The more the better and the least the better respectively but the demand may go against them.
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We can realized from analysis that 428124... could be the first 6 digits. A range from 9 to 18 is an option of length required for using extract-able 18 S.F. real numbers.
9 9 9 = 1 0 3 8 7 4 2 0 4 8 9 lo g 9 = 1 0 3 6 9 6 9 3 0 9 9 . 6 3 1 5 7 0 3 5 8 7 4 3 5 4 3 0 9 5 0 9 5 4 8 = 1 0 3 6 9 6 9 3 0 9 9 × 1 0 0 . 6 3 1 5 7 0 3 5 8 7 4 3 5 4 3 0 9 5 0 9 5 4 8
= 4 . 2 8 1 2 4 7 7 3 1 7 5 7 4 7 0 4 8 0 3 6 9 8 7 1 1 5 9 2 7 6 5 × 1 0 3 6 9 6 9 3 0 9 9
Scientific method which is useful is proven by mathematical method.
Answer: 4 2 8
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Let A = 9 9 9 ⟹ lo g 1 0 A = 9 9 lo g 1 0 9
By computation,
We have,
lo g 1 0 A = 9 9 lo g 1 0 9 = 3 . 6 9 6 9 3 0 9 9 6 3 1 5 7 0 3 5 8 7 4 3 5 4 3 0 9 5 0 9 5 4 8 × 1 0 8
Raising both sides to the power of 10,
⟹ A = 1 0 3 . 6 9 6 9 3 0 9 9 6 3 1 5 7 0 3 5 8 7 4 3 5 4 3 0 9 5 0 9 5 4 8 × 1 0 8
By computation,
So, A = 4 . 2 8 1 2 4 7 7 3 1 7 5 7 4 7 0 4 8 0 3 6 9 8 × 1 0 3 6 9 6 9 3 0 9 9